Let us first define Hypotenuse Leg (HL) congruence theorem:
<em>If the hypotenuse and one leg of a right angle are congruent to the hypotenuse and one leg of the another triangle, then the triangles are congruent.</em>
Given ACB and DFE are right triangles.
To prove ΔACB ≅ ΔDFE:
In ΔACB and ΔDFE,
AC ≅ DF (one side)
∠ACB ≅ ∠DFE (right angles)
AB ≅ DE (hypotenuse)
∴ ΔACB ≅ ΔDFE by HL theorem.
Y=7x-10
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Plz let me know if I am correct
It looks like the parabola are going down, so the coefficient should be negative,
B or C,
then we see that the parabola has x -intercepts (1,0) and (5,0)
these roots should be written as multiples
x=1, so x-1=0
and
x=5, so x-5=0,
so (x-1)(x-5) and negative first coefficient -2 at the same time.
We can see only in the
B. <span>f(x) = –2(x – 5)(x – 1)</span>
Just do 12 times 2 d=24 because 24 divided by 12 =2
Hello from MrBillDoesMath!
Answer:
40
Discussion:
A diagram is always appreciated!
Assuming that
mAOC = mAOB + mBOC =>
108 = (3x + 4) + (8x - 28) => combine common terms
108 = (3x + 8x) + (4 - 28 ) =>
108 = 11x - 24 => add 24 to both sides
132 = 11x =>
x = 132/11 = 12
So mAOB = 3x + 4 = 3(12) + 4 = 36 + 4 = 40
Thank you,
MrB
